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How can the following:

$$ \sqrt{27-10\sqrt{2}} $$

Be simplified to:

$$ 5 - \sqrt{2} $$


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7 Answers 7

up vote 20 down vote accepted

If you're faced with a question that says "Prove that $\sqrt{27-10\sqrt{2}}$ $=5 - \sqrt{2}$", then it's just a matter of squaring $5 - \sqrt{2}$ and seeing that you get $27-10\sqrt{2}$. But suppose the question your faced with is to find a square root of $27-10\sqrt{2}$ of the form $a+b\sqrt{2}$, where $a$ and $b$ are rational. Then you have $$ 27-10\sqrt{2}=\left(a+b\sqrt{2}\right)^2 = a^2 + 2ab\sqrt{2} + 2b^2 $$ so \begin{align} 27 & = a^2+2b^2 \\[8pt] -10 & = 2ab \end{align} From the second equation we get $a=-5/b$, then the first equation becomes $$ 27 = \frac{25}{b^2} + 2b^2 $$ or $$ 2(b^2)^2 -27b^2 + 25 = 0. $$ A solution is $b^2=1$, and you can go on from there to find $b$ and then $a$.

(And remember that the number will have two square roots.)

Later note: In order for all this to work, we have to rely on the fact that $\sqrt{2}$ is irrational. That enables us to conclude that the rational parts are equal and the irrational parts are equal, so we have two equations.

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It might be worth to mention that $5-\sqrt 2>0$. –  Git Gud Mar 13 '13 at 23:31

No lucky guesses are needed, there is a simple denesting algorithm for $\rm\:\sqrt{a+b\sqrt{n}}$

Simple Denesting Rule $\rm\ \ \ \color{#0A0}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

$\begin{array}{lll}\rm Recall\ \ \ w = a + b\sqrt{n}\ \ \ has\!\!\! &{\bf norm} &\!\!\!\rm=\: w\:\cdot\: w' = &\!\!\!\!\rm(a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 \\ \\ \rm and,\ furthermore,\ \ w\ \ has \!\!\!&{\bf trace} &\!\!\!\!\rm =\: w+w' = &\!\!\!\!\rm (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a\end{array}$

Here $\:27-10\sqrt{2}\:$ has norm $= 23^2.\,$ $\rm\, \color{#0A0}{Subtracting\ out}\ \sqrt{norm}\ = -23\ $ yields $\ 50-10\sqrt{2}\:$

and this has $\rm\ \sqrt{trace}\: =\: 10,\ \ hence\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields $\rm\ 5 - \sqrt{2} =\:$ sought sqrt.

Remark $\ $ The sign of the norm sqrt was chosen to make the trace sqrt rational. The same answer would arise using the opposite sign, but with slightly more work (rationalizing a denominator). $\ $ For many further examples see other posts on radical denesting.

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Note that $27=5^2+\sqrt2^2$ and $10\sqrt2=2\times5\times\sqrt2$

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Hint: Compare the squares of both expressions.

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Notice that $$\begin{align}27-10\sqrt{2} & = 25 - 2\cdot 5\sqrt{2} + 2 \\ & = 5^2 - 2\cdot5\sqrt2 + \left(\sqrt2\right)^2\\ & = \left(5 - \sqrt 2\right)^2 \end{align} $$

I don't know any way to notice this except to just get lucky and notice it; I didn't realize it could happen until sometime in high school when I was astounded to discover that $\sqrt{7+4\sqrt3} = 2+\sqrt3$.

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Set the nested radical as the difference of two square roots so that $$\sqrt{27-10\sqrt{2}}=(\sqrt{a}-\sqrt{b})$$ Then square both sides so that $$27-10\sqrt{2}=a-2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=27$$ and set (2) $$-2\sqrt{a}\sqrt{b}=-10\sqrt{2}$$ Square both sides of (2) to get $$4ab= 200$$ and solve for $b$ to get $$b=\frac{50}{a}$$ Replacing $b$ in (1) gives $$a+\frac{50}{a}=27$$ Multiply all terms by $a$ and convert to the quadratic equation $$a^2-27a+50=0$$ Solving the quadratic gives $a=25$ or $a=2$. Replacing $a$ and $b$ in the first difference of square roots formula above with $25$ and $2$ the solutions to the quadratic we have $$\sqrt{25}-\sqrt{2}$$ or $$5-\sqrt{2}$$

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$$\sqrt{27 - 10\sqrt{2}}$$ $$= \sqrt{25 + 2 - 2.5.\sqrt{2}}$$ $$= \sqrt{5^{2} - 2.5.\sqrt{2} +{\sqrt{2}}^2}$$ $$= \sqrt{{(5 - \sqrt{2})}^2}$$

$$= (5 - \sqrt{2})$$ $$= R.H.S$$

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You need equals signs rather than arrows. I would have written something similar as $$\sqrt{27-10\sqrt{2}} = \sqrt{27-2\sqrt{ 25 \times 2}} = \sqrt{25-2\sqrt{ 25 }\sqrt{2} +2} = \sqrt{(\sqrt{25} - \sqrt{2})^2}$$ –  Henry Mar 14 '13 at 8:51
@Henry: Thanks corrected. Its been ages I am solving some math problem –  Abhijit Mar 14 '13 at 9:13

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